Find the range of f(x) =`(sinx)/(x)+(x)/(tanx) in (0,(pi)/(2))`

Explanation

To find the range of the function \( f(x) = \frac{\sin x}{x} + \frac{x}{\tan x} \) for \( x \) in the interval \( (0, \frac{\pi}{2}) \), we will follow these steps:

### Step 1: Analyze the function
We start by rewriting the function:
\[ f(x) = \frac{\sin x}{x} + \frac{x}{\tan x} \]
We know that \( \tan x = \frac{\sin x}{\cos x} \), so we can express \( \frac{x}{\tan x} \) as:
\[ \frac{x}{\tan x} = \frac{x \cos x}{\sin x} \]
Thus, the function can be rewritten as:
\[ f(x) = \frac{\sin x}{x} + \frac{x \cos x}{\sin x} \]

### Step 2: Differentiate the function
To find the monotonicity of the function, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( \frac{\sin x}{x} \right) + \frac{d}{dx} \left( \frac{x \cos x}{\sin x} \right) \]
Using the quotient rule, we find:
\[ f'(x) = \frac{x \cos x - \sin x}{x^2} + \frac{\sin x \cdot \cos x - x \sin x \cdot \sin x}{\sin^2 x} \]

### Step 3: Analyze the sign of the derivative
We need to analyze the sign of \( f'(x) \) in the interval \( (0, \frac{\pi}{2}) \):
- The term \( \frac{x \cos x - \sin x}{x^2} \) is negative since \( \cos x \) decreases and \( \sin x \) increases in this interval.
- The term \( \frac{\sin x \cdot \cos x - x \sin^2 x}{\sin^2 x} \) is also negative because \( \sin x \cdot \cos x \) increases slower than \( x \cdot \sin^2 x \).

Since both parts of \( f'(x) \) are negative, we conclude that \( f'(x) < 0 \) for \( x \in (0, \frac{\pi}{2}) \). Therefore, \( f(x) \) is a decreasing function in this interval.

### Step 4: Evaluate the limits at the endpoints
Now, we evaluate the limits of \( f(x) \) as \( x \) approaches the boundaries of the interval:
1. As \( x \to 0 \):
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{\sin x}{x} + \frac{x}{\tan x} \right) = 1 + 1 = 2 \]
2. As \( x \to \frac{\pi}{2} \):
\[ \lim_{x \to \frac{\pi}{2}} f(x) = \frac{\sin(\frac{\pi}{2})}{\frac{\pi}{2}} + \frac{\frac{\pi}{2}}{\tan(\frac{\pi}{2})} = \frac{2}{\pi} + 0 = \frac{2}{\pi} \]

### Step 5: Determine the range
Since \( f(x) \) is decreasing from \( 2 \) to \( \frac{2}{\pi} \) as \( x \) moves from \( 0 \) to \( \frac{\pi}{2} \), the range of \( f(x) \) is:
\[ \left( \frac{2}{\pi}, 2 \right) \]

### Final Answer
The range of \( f(x) = \frac{\sin x}{x} + \frac{x}{\tan x} \) in the interval \( (0, \frac{\pi}{2}) \) is:
\[ \left( \frac{2}{\pi}, 2 \right) \]